I'm trying to use a dual quaternion Hand Eye Calibration Algorithm Header and Implementation, and I'm getting values that are way off. I'm using a robot arm and an optical tracker, aka camera, plus a fiducial attached to the end effector. In my case the camera is not on the hand, but instead sitting off to the side looking at the arm.

The transforms I have are:

  • Robot Base -> End Effector
  • Optical Tracker Base -> Fiducial

The transform I need is:

  • Fiducial -> End Effector


I'm moving the arm to a series of 36 points on a path (blue line), and near each point I'm taking a position (xyz) and orientation (angle axis with theta magnitude) of Camera->Fiducial and Base->EndEffector, and putting them in the vectors required by the HandEyeCalibration Algorithm. I also make sure to vary the orientation by about +-30 degrees or so in roll pitch yaw.

I then run estimateHandEyeScrew, and I get the following results, and as you can see the position is off by an order of magnitude.

[-0.0583, 0.0387, -0.0373] Real [-0.185, -0.404, -0.59] Estimated with HandEyeCalib

Here is the full transforms and debug output:

# INFO: Before refinement: H_12 =
-0.443021 -0.223478  -0.86821  0.321341
 0.856051 -0.393099 -0.335633  0.470857
-0.266286 -0.891925   0.36546   2.07762
        0         0         0         1
Ceres Solver Report: Iterations: 140, Initial cost: 2.128370e+03, Final cost: 6.715033e+00, Termination: FUNCTION_TOLERANCE.
# INFO: After refinement: H_12 =
  0.896005   0.154992  -0.416117  -0.185496
 -0.436281    0.13281  -0.889955  -0.404254
-0.0826716   0.978948   0.186618  -0.590227
         0          0          0          1

expected RobotTipToFiducial (simulation only):   0.168   -0.861    0.481  -0.0583
expected RobotTipToFiducial (simulation only):   0.461   -0.362    -0.81   0.0387
expected RobotTipToFiducial (simulation only):   0.871    0.358    0.336  -0.0373
expected RobotTipToFiducial (simulation only):       0        0        0        1

estimated RobotTipToFiducial:   0.896    0.155   -0.416   -0.185
estimated RobotTipToFiducial:  -0.436    0.133    -0.89   -0.404
estimated RobotTipToFiducial: -0.0827    0.979    0.187    -0.59
estimated RobotTipToFiducial:       0        0        0        1

Am I perhaps using it in the wrong way? Is there any advice you can give?

  • Are you trying to calibrate the transform from the base of the arm to the camera rig, or the kinematics of the arm, or something else? – Ben May 5 '15 at 17:54
  • I'm trying to calibrate between the end effector and the fiducial. – Andrew Hundt May 5 '15 at 19:48
  • The bug was that the first transform was not initialized properly – Andrew Hundt Jun 12 '15 at 2:30
up vote 7 down vote accepted

After solving the problem, I created a keynote presentation explaining many details about hand eye calibration for those that are interested. Practical code and instructions to calibrate your robot can be found at handeye-calib-camodocal.

I've directly reproduced some key aspects answering the question here.

Hand Eye Calibration Basics

Two Common Solutions to Hand Eye Calibration

AX=XB Hand Eye Calibration Solution

AX=ZB Hand Eye Calibration Solution


Camodocal is the library I'm using to solve the problem. It is a well written C++ library that includes hand eye calibration, though documentation is extremely sparse.

Camodocal includes implementations of:

Relevant Papers

  • CamOdoCal: Automatic Intrinsic and Extrinsic Calibration of a Rig with Multiple Generic Cameras and Odometry, In Proc. IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS), 2013.
  • Lionel Heng, Mathias Bürki, Gim Hee Lee, Paul Furgale, Roland Siegwart, and Marc Pollefeys, Infrastructure-Based Calibration of a Multi-Camera Rig, Submitted to IEEE International Conference on Robotics and Automation, 2014.

Feeding data into CamOdoCal

  1. Each measurement taken at a different time, position, and orientation narrows down the possible transforms that can represent the unknown X

  2. Record a list of many transforms A and B taken between different time steps, or relative to the first time step

    • Rotations are in AxisAngle = UnitAxisAngle format, or [x_axis,y_axis,z_axis]𝜃_angle
      • ||UnitAxis||=1
      • || AxisAngle || = 𝜃_angle
    • Translations are in the normal [x,y,z] format
  3. Pass both vectors into EstimateHandEyeScrew()
  4. Returns X in the form of a 4x4 transform estimate

Camodocal Hand Eye Calibration Details

ROS Integration

There is a package that implements a solver for this using camodocal + ROS called handeye_calib_camodocal. It also includes detailed troubleshooting instructions.


  • Strobl, K., & Hirzinger, G. (2006) . Optimal hand-eye calibration. In 2006 IEEE/RSJ international conference on intelligent robots and systems (pp. 4647–4653), October 2006.
  • Technical University of Munich (TUM) CAMP lab wiki

  • K. Daniilidis, “Hand–Eye Calibration Using Dual Quaternions,” Int. Journal of Robs. Research, vol. 18, no. 3, pp. 286–298, June 1999.
  • E. Bayro–Corrochano, K. Daniilidis, and G. Sommer, “Motor–Algebra for 3D Kinematics: The Case of Hand–Eye Calibration,” Journal for Mathem. Imaging and Vision, vol. 13, no. 2, pp. 79–100, Oct. 2000.
  • F. Dornaika and R. Horaud, “Simultaneous Robot–World and Hand– Eye Calibration,” IEEE Trans. on Robs. and Aut., vol. 14, no. 4, pp. 617–622, August 1998.
  • Note: figures and text are from mixed sources including the presentation author, the various papers referenced, and the TUM wiki.

The library you are using (and papers it is based on) seem to be for a different use case than what you are doing. They have a camera rig moving around a world with fiducials pulled out of a SLAM map, whereas you have a static camera and a moving arm holding fiducials. Fundamentally, yes they are the same, but i wonder if you used a different library that is more suited to a robot arm, if you would have better results. This is a paper you might consider: Calibrating a multi-arm multi-sensor robot: A Bundle Adjustment Approach. I don't know if there is source code. But it is ROS, so odds are good.

  • They are different applications, but from what I have read they seem to both be slightly different constructions of the same AX=XB problem. Here is the Daniilidis paper I believe the code is implementing, and here is another paper with some additional useful explanation. Perhaps I'm not understanding things correctly? – Andrew Hundt May 5 '15 at 19:44
  • I've been able to confirm that the problem is mathematically identical in my situation and in camodocal, even though the use case appears slightly different. – Andrew Hundt Jun 11 '15 at 23:25

I agree with Ben, your situation is eye-to-hand(camera outside robot) and AX=XB,the X is pose between camera and the robot base(but you get the fucucial with the end-effector). And in eye-in-hand(camera bind to robot tip) the X is the pose between camera and the robot's end-effector. There is another Matlab code (Dual Quaternions)in http://math.loyola.edu/~mili/Calibration/index.html

  • I also think that the X is pose between camera and the robot base for eye-to-hand. – 马继强 Dec 6 at 8:34

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